Questions: Unit 6 Homework Question 5 of 17 (1 point) I Question Attempt: 1 of 3 Assuming fixed amounts of gases, complete the following table by filling in any missing values corresponding to gases A, B, and C. Round each of your answers to 2 significant figures. Note: Reference the Conversion factors for non-SI units table for additional information. Gas P1 V1 T1 P2 V2 T2 --------------------- A 0.29 atm 4.34 L 340. K 0.84 atm 1.70 L 381. K B 1.1 atm 46.0 L 47. °C 574. mmHg 88. °C C 496. mmHg 329. mL 284. K 508. mmHg 0.81 L 450 K

Solution Steps

We are given initial and final states for gases A, B, and C. We need to find the missing values using the ideal gas law and the combined gas law.

For gas B, convert the pressure from mmHg to atm: \[ 574 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.7553 \text{ atm} \]

The combined gas law is: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Given: \[ P_1 = 1.1 \text{ atm}, \, V_1 = 46.0 \text{ L}, \, T_1 = 47^\circ \text{C} = 320 \text{ K} \] \[ P_2 = 0.7553 \text{ atm}, \, T_2 = 88^\circ \text{C} = 361 \text{ K} \]

Using the combined gas law to find \( V_2 \): \[ \frac{1.1 \times 46.0}{320} = \frac{0.7553 \times V_2}{361} \] \[ V_2 = \frac{1.1 \times 46.0 \times 361}{320 \times 0.7553} \approx 72.0 \text{ L} \]

Given: \[ P_1 = 496 \text{ mmHg}, \, V_1 = 329 \text{ mL}, \, T_1 = 284 \text{ K} \] \[ P_2 = 508 \text{ mmHg}, \, V_2 = 0.81 \text{ L} = 810 \text{ mL}, \, T_2 = 450 \text{ K} \]

Using the combined gas law to find \( T_2 \): \[ \frac{496 \times 329}{284} = \frac{508 \times 810}{T_2} \] \[ T_2 = \frac{508 \times 810 \times 284}{496 \times 329} \approx 450 \text{ K} \]

Final Answer